$f(x)=6{{x}^{4}}+2{{x}^{3}}$ What is the coefficient for the term containing $(x+1)^2$ in the Taylor polynomial, centered at $x=-1$, of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $42$ (Choice B) B $30$ (Choice C) C $20$ (Choice D) D $60$
Solution: Each term of a Taylor Series Polynomial centered at $~x=-1~$ is in the form of $\frac{{{f}^{(n)}}(-1){{(x+1)}^{n}}}{n!}$. We need the second derivative of $~f\left( x \right)~$ evaluated at $~x=-1\,$. $f(x)=6{{x}^{4}}+2{{x}^{3}}$ $f\,^\prime(x)=24{{x}^{3}}+6x^2$ $f\,^{\prime\prime}(x)=72x^2+12x$ So $f\,^{\prime\prime}(-1)=60\,$. Therefore, the coefficient of the term containing $~{{\left( x+1 \right)}^{2}}~$ is $~\frac{60}{2!}=30\,$.